Fie \( a \) , \( b\in (1,\infty) \) astfel încât \( ab=a+b \) . Să se rezolve ecuaţia :
\( x[(b-1)a^x+(a-1)b^x-x-ab+2]=(a^x-1)(b^x-1) \) .
Ecuatie din SHL 2009
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Ecuatia se scrie: \( x[(b-1)a^x+(a-1)b^x-x+(1-a)+(1-b)]=(a^x-1)(b^x-1) \)
\( \Longleftrightarrow\ x[(b-1)(a^x-1)+(a-1)(b^x-1)-x]=(a^x-1)(b^x-1) \)
\( \Longleftrightarrow\ x^2-x[(b-1)(a^x-1)+(a-1)(b^x-1)]+(a^x-1)(b^x-1)=0 \)
Avem : \( \Delta=[(b-1)(a^x-1)+(a-1)(b^x-1)]^2-4(a^x-1)(b^x-1) \) si cum \( (a-1)(b-1)=1 \)
\( \Longrightarrow\ \Delta=[(b-1)(a^x-1)-(a-1)(b^x-1)]^2 \) .
Prin urmare, \( x_{1,2}=\frac{(b-1)(a^x-1)+(a-1)(b^x-1)\ \pm\ |(b-1)(a^x-1)-(a-1)(b^x-1)|}{2} \)
\( \Longrightarrow\ \left{\ \begin{array}{ccc}
x_1=(b-1)(a^x-1) & = & \frac{a^x-1}{a-1} \\\\\\\\
x_2=(a-1)(b^x-1) & = & \frac{b^x-1}{b-1}\ \end{array} \) . Rezolvam prima ecuatie, a doua fiind identica .
\( x=\frac{a^x-1}{a-1}\ \Longleftrightarrow\ x(a-1)+1=a^x\ (\ast) \) . Insa \( f\ :\ \mathbb{R}\to\mathbb{R}\ ,\ f(x)=x(a-1)+1 \) este o functie liniara,
in timp ce \( f\ :\ \mathbb{R}\to\mathbb{R}\ ,\ a\ >\ 1\ ,\ f(x)=a^x \) este o functie convexa .
Deci ecuatia\( \ (\ast) \) are cel mult doua solutii . Cum \( x=0 \) si \( x=1 \) sunt solutii, inseamna ca ele sunt unice .
\( \Longleftrightarrow\ x[(b-1)(a^x-1)+(a-1)(b^x-1)-x]=(a^x-1)(b^x-1) \)
\( \Longleftrightarrow\ x^2-x[(b-1)(a^x-1)+(a-1)(b^x-1)]+(a^x-1)(b^x-1)=0 \)
Avem : \( \Delta=[(b-1)(a^x-1)+(a-1)(b^x-1)]^2-4(a^x-1)(b^x-1) \) si cum \( (a-1)(b-1)=1 \)
\( \Longrightarrow\ \Delta=[(b-1)(a^x-1)-(a-1)(b^x-1)]^2 \) .
Prin urmare, \( x_{1,2}=\frac{(b-1)(a^x-1)+(a-1)(b^x-1)\ \pm\ |(b-1)(a^x-1)-(a-1)(b^x-1)|}{2} \)
\( \Longrightarrow\ \left{\ \begin{array}{ccc}
x_1=(b-1)(a^x-1) & = & \frac{a^x-1}{a-1} \\\\\\\\
x_2=(a-1)(b^x-1) & = & \frac{b^x-1}{b-1}\ \end{array} \) . Rezolvam prima ecuatie, a doua fiind identica .
\( x=\frac{a^x-1}{a-1}\ \Longleftrightarrow\ x(a-1)+1=a^x\ (\ast) \) . Insa \( f\ :\ \mathbb{R}\to\mathbb{R}\ ,\ f(x)=x(a-1)+1 \) este o functie liniara,
in timp ce \( f\ :\ \mathbb{R}\to\mathbb{R}\ ,\ a\ >\ 1\ ,\ f(x)=a^x \) este o functie convexa .
Deci ecuatia\( \ (\ast) \) are cel mult doua solutii . Cum \( x=0 \) si \( x=1 \) sunt solutii, inseamna ca ele sunt unice .